Question

A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 rev/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.

Answer 1

Initial angular velocity of the wheel,

\[\omega = 100  rev/\min\] 

\[       = \frac{100}{60} = \frac{5}{3}\text{ rev/s }= \frac{10\pi}{3}\text{ rad/s}\] 

\[\theta = 10\text{ rev }= 20\pi\text{ rad}\] 

\[r = 0 . 2  m\]

Angular deceleration produced by the tangential force in order to stop the wheel after 10 revolutions,

\[\alpha = \frac{\omega^2}{2\theta}\]

Torque by which the wheel will come to rest,

\[\tau =  I_{cm} \alpha\]

\[\text{Or, }\Rightarrow   \tau = Fr =  I_{cm} \alpha\] 

Putting the values of `I_{cm}` and `alpha,` we  get

\[Fr = \frac{1}{2}m r^2  \times  \left( \frac{10\pi}{3} \right)^2  \times \frac{1}{2 \times 20\pi}\]

\[ \Rightarrow F = \frac{1}{2} \times 10 \times 0 . 2 \times \frac{100 \pi^2}{\left( 9 \times 2 \times 20\pi \right)}\]

\[ \Rightarrow F = \frac{5\pi}{18} = \frac{15 . 7}{18} = 0 . 87  N\]